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# Find $\sin \large\frac{x}{2},$$\cos \large\frac{x}{2} and \tan \large\frac{x}{2} in each of the following \tan x=-\large\frac{4}{3},x in quadrant II \begin{array}{1 1} \sin \frac{x}{2} =\frac{2}{\sqrt 5} ; \cos \frac{x}{2} =\frac{-1}{\sqrt 5} \\ \sin \frac{x}{2} =\frac{-2}{\sqrt 5} ; \cos \frac{x}{2} =\frac{1}{\sqrt 5} \\ \sin \frac{x}{2} =\frac{-2}{\sqrt 5} ; \cos \frac{x}{2} =\frac{-1}{\sqrt 5} \\ \sin \frac{x}{2} =\frac{2}{\sqrt 5} ; \cos \frac{x}{2} =\frac{1}{\sqrt 5} \end{array} Can you answer this question? ## 1 Answer 0 votes Toolbox: • \cos x=\large\frac{-1}{\sqrt{1\tan^2x}} Since x lies in the second quadrant \therefore \cos x is negative. \therefore \cos x=\large\frac{-1}{\sqrt{1+\tan^2x}}=-\frac{1}{\sqrt{1+\Large\frac{16}{9}}}=-\frac{3}{5} Now ,x lies in II quadrant \Rightarrow \large\frac{\pi}{2}$$ < x < \pi\Rightarrow \large\frac{\pi}{4} $$< \large\frac{x}{2}$$ < \large\frac{\pi}{2}$
$\Rightarrow \large\frac{x}{2}$ lies in first quadrant.
$\Rightarrow \sin \large\frac{x}{2}$$,\cos \large\frac{x}{2} and \tan \large\frac{x}{2} are positive \therefore\cos\large\frac{x}{2}=\sqrt{\large\frac{1+\cos x}{2}}=\sqrt{\large\frac{1-\Large\frac{3}{5}}{2}}=\sqrt{\large\frac{5-3}{2\times 5}}=\frac{1}{\sqrt 5} \therefore\sin\large\frac{x}{2}=\sqrt{\large\frac{1-\cos x}{2}}=\sqrt{\large\frac{1+\Large\frac{3}{5}}{2}}=\sqrt{\large\frac{5+3}{2\times 5}}=\frac{2}{\sqrt 5} \therefore\tan\large\frac{x}{2}=\sqrt{\large\frac{1-\cos x}{1+\cos x}}=\sqrt{\large\frac{1+\Large\frac{3}{5}}{1-\Large\frac{3}{5}}}=\sqrt{\large\frac{5+3}{5-3}}=$$\sqrt{4}=2$